Problem: Simplify the following expression and state the condition under which the simplification is valid. $z = \dfrac{4a^3 + 56a^2 + 180a}{a^3 + 10a^2 + 9a}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ z = \dfrac {4a(a^2 + 14a + 45)} {a(a^2 + 10a + 9)} $ $ z = \dfrac{4a}{a} \cdot \dfrac{a^2 + 14a + 45}{a^2 + 10a + 9} $ Simplify: $ z = 4 \cdot \dfrac{a^2 + 14a + 45}{a^2 + 10a + 9}$ Since we are dividing by $a$ , we must remember that $a \neq 0$ Next factor the numerator and denominator. $ z = 4 \cdot \dfrac{(a + 9)(a + 5)}{(a + 9)(a + 1)}$ Assuming $a \neq -9$ , we can cancel the $a + 9$ $ z = 4 \cdot \dfrac{a + 5}{a + 1}$ Therefore: $ z = \dfrac{ 4(a + 5)}{ a + 1 }$, $a \neq -9$, $a \neq 0$